Problem: Divide the following complex numbers. $ \dfrac{-3-15i}{-2+3i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2-3i}$ $ \dfrac{-3-15i}{-2+3i} = \dfrac{-3-15i}{-2+3i} \cdot \dfrac{{-2-3i}}{{-2-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-3-15i) \cdot (-2-3i)} {(-2+3i) \cdot (-2-3i)} = \dfrac{(-3-15i) \cdot (-2-3i)} {(-2)^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-3-15i) \cdot (-2-3i)} {(-2)^2 - (3i)^2} = $ $ \dfrac{(-3-15i) \cdot (-2-3i)} {4 + 9} = $ $ \dfrac{(-3-15i) \cdot (-2-3i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-3-15i}) \cdot ({-2-3i})} {13} = $ $ \dfrac{{-3} \cdot {(-2)} + {-15} \cdot {(-2) i} + {-3} \cdot {-3 i} + {-15} \cdot {-3 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{6 + 30i + 9i + 45 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{6 + 30i + 9i - 45} {13} = \dfrac{-39 + 39i} {13} = -3+3i $